Question: $g(t) = t-1+4(f(t))$ $f(n) = -3n^{2}-4n$ $h(x) = -4x+3-4(f(x))$ $ h(g(0)) = {?} $
First, let's solve for the value of the inner function, $g(0)$ . Then we'll know what to plug into the outer function. $g(0) = -1+4(f(0))$ To solve for the value of $g$ , we need to solve for the value of $f(0)$ $f(0) = -3(0^{2})+(-4)(0)$ $f(0) = 0$ That means $g(0) = -1+(4)(0)$ $g(0) = -1$ Now we know that $g(0) = -1$ . Let's solve for $h(g(0))$ , which is $h(-1)$ $h(-1) = (-4)(-1)+3-4(f(-1))$ To solve for the value of $h$ , we need to solve for the value of $f(-1)$ $f(-1) = -3(-1)^{2}+(-4)(-1)$ $f(-1) = 1$ That means $h(-1) = (-4)(-1)+3+(-4)(1)$ $h(-1) = 3$